IDNLearn.com helps you find the answers you need quickly and efficiently. Join our knowledgeable community and access a wealth of reliable answers to your most pressing questions.
Sagot :
To understand the geometric effect of the transformation matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
on vectors and areas in a 2-dimensional space, let’s analyze how it transforms some basic vectors and how it affects areas.
### 1. Effect on Unit Vectors
#### a. Unit Vector along the x-axis:
The unit vector along the x-axis is
[tex]$ \left[\begin{array}{c} 1 \\ 0 \end{array}\right] $[/tex]
When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 1 \\ 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 1 + 0 \cdot 0 \\ 0 \cdot 1 + \frac{1}{2} \cdot 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right] $[/tex]
Therefore, the unit vector along the x-axis is scaled down to
[tex]$\left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right]$[/tex].
#### b. Unit Vector along the y-axis:
The unit vector along the y-axis is
[tex]$ \left[\begin{array}{c} 0 \\ 1 \end{array}\right] $[/tex]
When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 0 \\ 1 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 0 + 0 \cdot 1 \\ 0 \cdot 0 + \frac{1}{2} \cdot 1 \end{array}\right] = \left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right] $[/tex]
Therefore, the unit vector along the y-axis is scaled down to
[tex]$\left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right]$[/tex].
### 2. Effect on Area
The area effect of a transformation matrix is determined by its determinant. The determinant of the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is:
[tex]$ \det \left( \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \right) = \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right) - 0 \cdot 0 = \frac{1}{4} $[/tex]
This value tells us that any area in the 2D space will be scaled by a factor determined by the determinant of the matrix. In this case, areas are scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex].
### Conclusion
The geometric effect of the transformation represented by the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is to scale both the x and y components of any vector by [tex]\( \frac{1}{2} \)[/tex]. This results in vectors being halved in length along both axes. Additionally, any area in the plane is scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex], as indicated by the determinant of the transformation matrix.
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
on vectors and areas in a 2-dimensional space, let’s analyze how it transforms some basic vectors and how it affects areas.
### 1. Effect on Unit Vectors
#### a. Unit Vector along the x-axis:
The unit vector along the x-axis is
[tex]$ \left[\begin{array}{c} 1 \\ 0 \end{array}\right] $[/tex]
When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 1 \\ 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 1 + 0 \cdot 0 \\ 0 \cdot 1 + \frac{1}{2} \cdot 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right] $[/tex]
Therefore, the unit vector along the x-axis is scaled down to
[tex]$\left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right]$[/tex].
#### b. Unit Vector along the y-axis:
The unit vector along the y-axis is
[tex]$ \left[\begin{array}{c} 0 \\ 1 \end{array}\right] $[/tex]
When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 0 \\ 1 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 0 + 0 \cdot 1 \\ 0 \cdot 0 + \frac{1}{2} \cdot 1 \end{array}\right] = \left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right] $[/tex]
Therefore, the unit vector along the y-axis is scaled down to
[tex]$\left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right]$[/tex].
### 2. Effect on Area
The area effect of a transformation matrix is determined by its determinant. The determinant of the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is:
[tex]$ \det \left( \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \right) = \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right) - 0 \cdot 0 = \frac{1}{4} $[/tex]
This value tells us that any area in the 2D space will be scaled by a factor determined by the determinant of the matrix. In this case, areas are scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex].
### Conclusion
The geometric effect of the transformation represented by the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is to scale both the x and y components of any vector by [tex]\( \frac{1}{2} \)[/tex]. This results in vectors being halved in length along both axes. Additionally, any area in the plane is scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex], as indicated by the determinant of the transformation matrix.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.
